TITLE: A particularly tricky integral: $ \lim_{n \to \infty} n^{\frac{3}{2}} \int_0^1 \frac{x^2}{(1+x^2)^n} dx. $
QUESTION [6 upvotes]: I encountered the following, deceptively simple looking integral - 
$$ \lim_{n \to \infty} n^{\frac{3}{2}} \int_0^1 \frac{x^2}{(1+x^2)^n} dx. $$
My goal was to hopefully somehow use the Lebesgue dominated convergence theorem; I've tried every elementary bound that I can think of to no avail. One approach suggested to me is to write 
$$ \int_0^1 \frac{x^2}{(1+x^2)^{n+1}}dx = \int_0^1 \frac{1}{(1+x^2)^n} - \int_0^1 \frac{1}{(1+x^2)^{n+1}}dx, $$
then try and use an induction and partial fraction decomposition argument. Doing this, I didn't find anything useful.
When I tried plugging into wolfram/mathematica I got roughly the behavior (this could be wrong, so take it with a grain of salt) 
$$\int_0^1 \frac{x^2}{(1+x^2)^n} dx =  {}_2 F_1\left(\frac{1}{2}, n, \frac{3}{2}, -1\right). $$
Other calculations by trying to play with Gaussian bounds led to the result $ \frac{\sqrt{\pi}}{4} $ but this was done on a computer, not with an explicit method. There's also been some hints from mathematica that $$ \lim_{n \to \infty} n^{\frac{3}{2}} \int_0^1 \frac{x^2}{(1+x^2)^n} dx $$ is somehow related to $$ \sqrt{\pi}\frac{\Gamma\left(n-\frac{1}{2}\right)}{2\Gamma(n)}. $$
I believe that there is an elementary argument and that I just haven't found it. Any tips would be appreciated.

REPLY [10 votes]: By the change of variable $y = n^{1/2} x$ you get
$$
n^{3/2} ∫_0^1\frac{x^2}{(1+x^2)^{n}}\,\mathrm{d}x = ∫_0^{n^{1/2}}\frac{y^2}{(1+n^{-1}y^2)^{n}}\,\mathrm{d}y \underset{n\to\infty}{⟶} ∫_0^\infty y^2e^{-y^2}\,\mathrm{d}y
$$
by dominated convergence. Now, do the change of variable $x = y^2$ to get
$$
∫_0^\infty y^2e^{-y^2}\,\mathrm{d}y = \frac{1}{2}∫_0^\infty x^{1/2} e^{-x}\,\mathrm{d}x = \frac{\Gamma(3/2)}{2} = \frac{\Gamma(1/2)}{4} = \frac{\sqrt{\pi}}{4}
$$